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azamat
3 years ago
14

Jennifer cut a length of rope into two pieces. One piece was 2 feet 9 inches long, and the other piece was 3 feet 5 inches long.

How long was the rope before it was cut?
A)5 feet 2 inches.
B) 5 feet 4 inches
C)6 feet 2 inches
D)6 feet 4 inches
E)6 feet 6 inches
Mathematics
1 answer:
GenaCL600 [577]3 years ago
6 0

Answer:

6ft 2ins

Step-by-step explanation:

ins:9+5=14

ft:2+3=5

now you can add another foot because there is 12 ins in 1 foot making the answer 6 feet and 2 inches

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How many times will 12 go into 6839
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56.916... reapeating

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Determine the value of cos 79º 15' (cosine of 79 degrees, and 15 minutes). In this case, minutes are 1/60 of a degree.
yKpoI14uk [10]

Answer:

A. 0.186524036

Step-by-step explanation:

Cosine 79° 15'

But, 1° = 60'

Thus; 15' = 15/60 = 0.25°

Therefore;

cos 79° 15' = cos 79.25°

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8 0
3 years ago
Of 580580 samples of seafood purchased from various kinds of food stores in different regions of a country and genetically compa
Lubov Fominskaja [6]

Answer:

a) The 99% confidence interval would be given (0.204;0.296).

b) We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

c) No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

Step-by-step explanation:

Part a

Data given and notation  

n=580 represent the random sample taken    

X represent the seafood sold in the country that is mislabeled or misidentified by the people

\hat p=0.25 estimated proportion of seafood sold in the country that is mislabeled or misidentified by the people

\alpha=0.01 represent the significance level (no given, but is assumed)    

p= population proportion of seafood sold in the country that is mislabeled or misidentified by the people

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

0.25 - 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.204

0.25 + 2.58 \sqrt{\frac{0.25(1-0.25)}{580}}=0.296

And the 99% confidence interval would be given (0.204;0.296).

Part b

We have 99% of confidence that the true population proportion of all seafood sold in the country that is mislabeled or misidentified is between (0.204;0.296).  

Part c

A government spokesperson claimed that the sample size was too​ small, relative to the billions of pieces of seafood sold each​ year, to generalize. Is this criticism​ valid?

No that's not true. Because the necessary assumptions and conditions for the confidence interval for the proportion are satisifed, so then we can use inferential statistics to interpret the interval to the population of interest.

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Line 4. 12 would be divided by 6, not multiplied. 
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